Terminal velocity is the maximum velocity attainable by an object as it falls through a fluid and is given by,
Where,
m = mass of the falling object.
g = the acceleration due to gravity.
\rho = the density of the fluid the object is falling through.
A = the projected area of the object.
C = the drag coefficient.
A) Because a drop is difficult to approximate to a certain form, it usually tends to be considered a spectrum, the terminal velocity for a sphere is given by
Whatever our appreciations, all the variables are constant, except for the Diameter, we can realize that the terminal velocity is proportional to the radius of the object, the greater the radius - the larger the drop - the greater the terminal velocity.
B) Since there is a "constant" terminal velocity at the end of the path, at which point the forces are balanced, the acceleration will be 0. For both objects.
I would say that sand in water is a suspension as you can see the particles with the naked eye or a microscope. Also, the solution will settle if left alone and the sand will sink to the bottom.
The power of the machine is 50
Answer:
a) They are in the same point
b) t = 0 s, t = 2.27 s, t = 5.73 s
c) t = 1 s, t = 4.33 s
d) t = 2.67 s
Explanation:
Given equations are:
Constants are:
a) "Just after leaving the starting point" means that t = 0. So, if we look the equations, both and depend on t and don't have constant terms.
So both cars A and B are in the same point.
b) Firstly, they are in the same point in x = 0 at t = 0. But for generalized case, we must equalize equations and solve quadratic equation where roots will give us proper t value(s).
s, s
c) Since the distance isn't changing, the velocities are equal. To find velocities, we need to take the derivatives of both equations with respect to time and equalize them.
s, s
d) For same acceleration, we we need to take the derivatives of velocity equations with respect to time and equalize them.
s
Answer:
v_{ average} = 5.57
Explanation:
The most probable value of a measure is
v_average = ∑ x_i
where N is the number of measurements
in tes case N = 3
v_{average} = ⅓ (5.63 +5.54 + 5.53)
V_{average} = 5,567
The number of significant figures must be equal to the number of figures that have the least in the readings.
v_{ average} = 5.57